Find the limit of the sequence 22 – 2 32 – 2 42 – 2 52 – 2 - 22 32 42 52
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Find the Limit of the Sequence:**
The sequence is given by:
\[
\frac{2^2 - 2}{2^2}, \quad \frac{3^2 - 2}{3^2}, \quad \frac{4^2 - 2}{4^2}, \quad \frac{5^2 - 2}{5^2}, \ldots
\]
**Explanation:**
- The sequence is in the form \(\frac{n^2 - 2}{n^2}\), where \(n\) is a positive integer starting from 2.
- For the first term, \(n = 2\):
\[
\frac{2^2 - 2}{2^2} = \frac{4 - 2}{4} = \frac{2}{4} = \frac{1}{2}
\]
- For the second term, \(n = 3\):
\[
\frac{3^2 - 2}{3^2} = \frac{9 - 2}{9} = \frac{7}{9}
\]
- For the third term, \(n = 4\):
\[
\frac{4^2 - 2}{4^2} = \frac{16 - 2}{16} = \frac{14}{16} = \frac{7}{8}
\]
- For the fourth term, \(n = 5\):
\[
\frac{5^2 - 2}{5^2} = \frac{25 - 2}{25} = \frac{23}{25}
\]
The pattern shows each term is of the form \(\frac{n^2 - 2}{n^2} = 1 - \frac{2}{n^2}\).
**Limit Calculation:**
\[
\lim_{n \to \infty} \left(1 - \frac{2}{n^2}\right) = 1 - 0 = 1
\]
Therefore, the limit of the sequence as \(n\) approaches infinity is 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14ad990f-dd7d-47e0-a13f-1f780fa26f58%2F4d610fcc-fd9e-4353-9870-47aa60abb909%2Fnv8gtn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Find the Limit of the Sequence:**
The sequence is given by:
\[
\frac{2^2 - 2}{2^2}, \quad \frac{3^2 - 2}{3^2}, \quad \frac{4^2 - 2}{4^2}, \quad \frac{5^2 - 2}{5^2}, \ldots
\]
**Explanation:**
- The sequence is in the form \(\frac{n^2 - 2}{n^2}\), where \(n\) is a positive integer starting from 2.
- For the first term, \(n = 2\):
\[
\frac{2^2 - 2}{2^2} = \frac{4 - 2}{4} = \frac{2}{4} = \frac{1}{2}
\]
- For the second term, \(n = 3\):
\[
\frac{3^2 - 2}{3^2} = \frac{9 - 2}{9} = \frac{7}{9}
\]
- For the third term, \(n = 4\):
\[
\frac{4^2 - 2}{4^2} = \frac{16 - 2}{16} = \frac{14}{16} = \frac{7}{8}
\]
- For the fourth term, \(n = 5\):
\[
\frac{5^2 - 2}{5^2} = \frac{25 - 2}{25} = \frac{23}{25}
\]
The pattern shows each term is of the form \(\frac{n^2 - 2}{n^2} = 1 - \frac{2}{n^2}\).
**Limit Calculation:**
\[
\lim_{n \to \infty} \left(1 - \frac{2}{n^2}\right) = 1 - 0 = 1
\]
Therefore, the limit of the sequence as \(n\) approaches infinity is 1.
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